时间限制

3 1
000007 James 85
000010 Amy 90
000001 Zoe 60

000001 Zoe 60
000007 James 85
000010 Amy 90

4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98

000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60

4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90

000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90

============================================
这个题目能想到最好的办法，就是用set然后重载一个自定义排序了 感谢郭嘉，感谢STL！
set这个东西嘛，好像是用红黑树还是什么实现的(印象中是这样的) 原来打算用vector的，但是vector好像不能在插入的时候就根据自定义的排序方法排序好插进去，这样如果全部录入完了整体排序一遍是不是更耗费时间呢？
参考文章：
http://blog.csdn.net/stone_sky/article/details/8471722 讲的是自定义排序 http://zhidao.baidu.com/question/189798009.html 讲的是输出填0格式化问题 http://www.189works.com/article-43335-1.html http://wenku.baidu.com/view/08c0eb0bff00bed5b9f31d29.html 这两个就是自定义排序了
这题目这么搞了以后就简单多了:)当然最后一个测试点还是用了100ms…不管了过了就行 ============================================

#include
#include
#include
#include
using namespace std;
int compareName(const char name1[],const  char name2[])
{
    for(int i=0; i<8; i++)
    {
        if( name1[i] < name2[i] )
            return 1;
        else if(name1[i] > name2[i])
            return -1;
        //相等的话继续比较
    }
    return 0; //相同的名字
}
struct StudentRecord
{
    int ID;
    char name[9];
    int grade;
    int sortType; //按哪一列排序，1,2,3
    StudentRecord(){}
    StudentRecord(int _id, char _name[], int _grade, int _sortType)
    {
        ID = _id;
        strcpy(name, _name);
        grade = _grade;
        sortType = _sortType;
    }
    void set(int _id, char _name[], int _grade)
    {
        ID = _id;
        strcpy(name, _name);
        grade = _grade;
    }
    //自定义排序方法
    bool operator< (const StudentRecord& other) const
    {
        int cmpResult;
        switch(sortType)
        {
        case 1:
            //按照ID升序排列
            return ID < other.ID;
        case 2:
            //按照姓名的升序排列，如果有同名再按ID升序
            cmpResult = compareName(name, other.name);
            if(cmpResult > 0)
                return true;
            else if (cmpResult < 0)
                return false;
            else
                return ID < other.ID;
        case 3:
            //按照成绩的升序排列，如果有同名成绩按ID升序
            if (grade == other.grade)
                return ID < other.ID;
            else
                return grade < other.grade;
        default:
            printf("ERROR!");
            return false;
        }
    }
};
set records;
int main()
{
    long N; int C;
    scanf("%ld %d", &N, &C);
    StudentRecord singleRec;
    singleRec.sortType = C;
    for(long i=0; i<N; i++)
    {
        int ID, grade;
        char name[8];
        scanf("%d %s %d", &ID, name, &grade);
        singleRec.set(ID, name, grade);
        records.insert(singleRec);
    }
    set::iterator it;
    for(it = records.begin(); it != records.end(); it++)
    {
        StudentRecord sr = *it;
        printf("%06d %s %dn", sr.ID, sr.name, sr.grade);
    }
    return 0;
}

====================================

测试点

时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
This time, you are supposed to find A*B where A and B are two polynomials.
Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a_N1 N2 a_N2 … NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < … < N2 < N1 <=1000.
 
Output Specification:
For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.
Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

=====================================
这题目没什么难度的说，就是多项式乘积嘛~
唯一要注意的事情是，结果的输出时按照指数递减的…这里用了map里面的reverse_iterator，真心是好用！
另外，也不知道系数有没有负数，所以处理的时候考虑了一下相加为0的情况…貌似没啥必要哦…
总之是很简单的东西~前提是有STL这货…
=====================================

#include 
#include